Download Extension worksheet – Topic 6 - Cambridge Resources for the IB

Cambridge Physics for the IB Diploma
Answers to Coursebook questions – Chapter 5.2
1
a
See Figure 2.4 on page 291 in Physics for the IB Diploma.
b
See Figure 2.6 on page 291 in Physics for the IB Diploma.
c
Something like Figure 2.6 but the place in the middle where the lines appear to
be making a square should be moved a bit to the left.
d
Essentially Figure 2.5 on page 291 in Physics for the IB Diploma (it does not
matter that we say three times in Physics for the IB Diploma that the difference
between 2 and 3 cannot really be seen).
2
Because if they were not there would a component of the electric field along the
surface. This would cause electrons to move in the conducting surface contrary to the
assumption that we are dealing with electrostatics in which charges cannot move.
3
a
b
4
a–c The force is the same everywhere inside the plates since the electric field is
V 500
uniform and equal to E  
 5.00 103 V m 1 .
d 0.10
The force is everywhere F  eE  1.6 1019  5.00 103  8.0 1016 N .
d
500
 50 volt for every cm moved up from the bottom
10
plate. So the potential difference between the two given points is 300 V. Hence
the work done is W  eV  1.6 1019  300  4.8 1017 J .
The potential increases by
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5
The force is F  qE  1.6  1019  100  1.6  1017 N and the acceleration is
a
6
7
F 1.6  10 17

 1.8  1013 ms2 .
31
m 9.1  10
F 3.0 105
F  eE  E  
 6.0 N C1 .
6
e 5.0 10
The magnitude of each of the fields produced at P is:
kQ 8.99 109  2.00 106
E 2 
 1.95 105 N C1 .
2
r
0.052  0.302


The vertical components of the electric fields will cancel out, leaving only the
horizontal components.
The horizontal component is
d
0.30
Ex  E cos   E
 1.95 105 
 1.92 105 N C1 .
2
2
a
0.10
d2 
 0.302
4
4
The net field is then directed horizontally to the right and has magnitude
2 1.92 105  3.84 105 N C1 .
8
The 2 electric fields are E1  1.95  105 NC1 and E2  3.90  10 5 NC1 .
Adding vectorially by taking components gives
Ex  1.92  105 NC1  3.85  105 NC1  5.77  105 NC1 and
Ey  0.3156 105 NC1  0.6313 105 NC1  0.3157 105 NC1 .
And so E  5.78  105 NC1 ,   arctan
9
10
0.3157  10 5
 3.1o .
5.77  10 5
a
W  qV  5.0 100  500 V .
b
W  qV  5.0  200  1000 V .
c
It would be the same, but only the initial and final points count, not the path
joining them.
a
Using eV 
2eV
2 1.6 1019 100
1 2
mv we find v 

 5.9 106 m s 1 .
2
m
9.11031
b
v
11
2eV
2 1.6 1019 100

 1.4 105 m s 1
27
m
1.67 10
The net electric field has magnitude E  1152  1252  170 N C1 .
115
Its direction is   arctan
 180  42.6  137.4 .
125
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12
a and b These points are inside the conducting sphere, so the electric field is
zero there.
c
E
d
kQ 8.99 109  4.0 106

 1.6 106 N C1
2
2 2
R
(15.0 10 )
At 20 cm.
2
 15 
E  1.6  10 6     9.0  10 5 NC1
 20 
13
14
a
The electric force is downwards, i.e. in the direction of the field, and so the
charge has to be positive.
b
Before the field is turned on, kx0  mg . With the field on, k2x0  mg  qE .
mg
Hence, 2mg  mg  qE and so E 
.
q
c
Calling the extension from equilibrium x we have that the net force on the
particle is Fnet  k(2x0  x)  qE  mg  kx . The force is upward, i.e. opposite the
displacement and proportional to displacement. Hence SHM oscillations take
place.
d
The net upward force is the same as what we would have without the electric
field, so the period is unaffected by its presence.
a
The electric field is E 
b
When it is displaced to the right, the force will be directed to the left and so the
mass will move back. It will overshoot the equilibrium position of a and move
past it to the left. Once there, the net force will now be directed to the right and so
the mass will keep oscillating.
c
To decide if the oscillations are SHM we must find the net force on the charge
and see that it is opposite and proportional to the displacement: let x be the
displacement to the right (measured from the position of equilibrium in a). The
kQq
9kQq
net force is F  qE 
which is directed to the left but is not

d
3d
2
2
(  x)
(  x)
4
4
proportional to x, so the oscillations are not SHM.
d
Displacing a negative charge to the right results in a net force to the right, i.e.
there will not be any oscillations.
kQ
k9Q

 0.
2
(d / 4) (3d / 4)2
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15
a
The net force has been found in Q14 to be F 
b
The force can be rewritten as
kQq
9kQq
.

d
3d
(  x)2 (  x)2
4
4


kQq
9kQq
16kQq 
1
1

F 2
 2


2

.
4x
4x
d
4x 2 9d
4x 2
d
2
2
 (1  ) (1  ) 
(1  )
(1  )
d
3d
16
d
16
3d
Applying the given approximation,
F
c
16kQq 
4x
4x  16kQq 8x 8x
256kQq
(
 )
x.
 (1  2 )  (1  2 ) 
2
2
d
d
3d
d
d 3d
3d 3
The force is opposite and proportional to the displacement so we will have SHM.
256kQq
256kQq
We have for the acceleration a  
.
x and so  2 
3
3md
3md 3
2π
The period will be T 
.

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